Question

A continuous function f : R  R satisfies the differential equation $$f\left(x\right)=(2+x^2)[3+\underset{0}{\overset{x}{\int }}\frac{f^2\left(t\right)}{2+t^2}dt]$$, then the value of f(2) is equal to

• A. $\frac{18}{19}$
• B. $\frac{17}{19}$
• C. $\frac{-18}{19}$
• D. $\frac{-17}{19}$

Answer 1

The correct option is C $\frac{-18}{19}$

From given differential equation
$\frac{f\left(x\right)}{2+x^2}=3+\underset{0}{\overset{x}{\int }}\frac{f^2\left(t\right)dt}{2+t^2}$ ,
where $f\left(x\right)=2[3+0]=6$
Differentiate both sides w.r.t. $x$
$\frac{2+x^2{f}^{\prime }\left(x\right)-2xf\left(x\right)}{{(2+x^2)}^2}=0+\frac{f^2\left(x\right)}{2+x^2}$
$\Rightarrow f^{\prime }\left(x\right)-\left(\frac{2x}{2+x^2}\right)f\left(x\right)=f^2\left(x\right)$
$\Rightarrow \frac{dy}{dx}-\left(\frac{2x}{2+x^2}\right)\cdot y=y^2$
$\Rightarrow \frac{1}{y^2}\frac{dy}{dx}-\left(\frac{2x}{2+x^2}\right)\cdot \frac{1}{y}=1$ ...(i)
Put
$-\frac{1}{y}=z\Rightarrow \frac{1}{y^2}\frac{dy}{dx}=\frac{dz}{dx}$
$\therefore$ Equation (i) becomes
$\frac{dz}{dx}+\left(\frac{2x}{2+x^2}\right)z=1$ ...(ii)
I.F.
$=e^{\int \frac{2x}{2+x^2}dx}=e^{{\log }_e(2+x^2)}=2+x^2$
$\therefore$ Solution of equation (ii) is
$\therefore z\cdot (2+x^2)=\int (2+x^2)dx$
$\Rightarrow z(2+x^2)=2x+\frac{x^3}{3}+c$
$\Rightarrow -\frac{1}{y}(2+x^2)=2x+\frac{x^3}{3}+c$
where $x=0$, $-\frac{1}{y_{\left(0\right)}}(2+0)=0+0+c$
$\Rightarrow c=-\frac{2}{6}=-\frac{1}{3}$
$\therefore -\frac{1}{y}(2+x^2)=2x+\frac{x^3}{3}-\frac{1}{3}$
$\Rightarrow \frac{(2+x^2)}{y}=\frac{6x+x^3-1}{3}$
$\Rightarrow y=f\left(x\right)=\frac{-3(2+x^2)}{x^3+6x-1}$
$\therefore f\left(2\right)=\frac{-3(2+4)}{8+12-1}=\frac{-18}{19}$